(\({\cal A}_{{\bf q}}\!\!\equiv \!{\cal A}_{-{\bf q}}^{\dag },\;\)  \({\cal B}_{\bf q}^{\dag }\)  means the \(a\rightarrow b\)  substitution; we note that \({\cal B}_{\bf q}^{\dag }|0\rangle \!\equiv \!0\)  and \({\cal A}_{\bf q}^{\dag }|0\rangle \!\equiv \!\delta _{{\bf q},0}|0\rangle \) ). When so doing, comparing to previous works,[1]}, [2]} one has taken into account that the \(a_p\)  and \(b_p\)  operators belong to different Landau levels. Therefore, the factors in the square brackets at \({\cal A}_{{\bf q}}\)  and \({\cal B}_{{\bf q}}\)  in Eq. (REF ) turn out to be Laguerre polynomials \(L_0(q^2\!/2)\!\equiv \!1\)  and \(L_1(q^2\!/2)\!\equiv \!(1\!-\!q^2\!/2)\) .
